Advertisements
Advertisements
Question
Factorise:
8p3 −\[\frac{27}{p^3}\]
Sum
Advertisements
Solution
It is known that,
a3 − b3 = (a − b)(a2 + ab + b2)
\[\ 8p^3 - \frac{27}{p^3}\]
\[ = \left(2p \right)^3 - \left(\frac{3}{p}\right)^3\]
\[ = \left(2p - \frac{3}{p} \right)\left\{\left(2p \right)^2 + \left( \frac{3}{p} \right)^2 + \left(2p \right) \times \left(\frac{3}{p} \right) \right\}\]
\[ = \left(2p - \frac{3}{p} \right)\left(4 p^2 + \frac{9}{p^2} + 6 \right)\]
shaalaa.com
Is there an error in this question or solution?
RELATED QUESTIONS
Simplify:
\[\frac{4 x^2 - 11x + 6}{16 x^2 - 9}\]
Simplify:
\[\frac{a^3 - 27}{5 a^2 - 16a + 3} \div \frac{a^2 + 3a + 9}{25 a^2 - 1}\]
Factorise:
y3 − 27
Factorise:
27m3 − 216n3
Factorise:
125y3 − 1
Simplify:
(3a + 5b)3 − (3a − 5b)3
Simplify:
p3 − (p + 1)3
Factorise: 54p3 - 250q3.
Simplify: (a - b)3 - (a3 - b3)
Simplify: (2x + 3y)3 - (2x - 3y)3
